Homework Help:
Lagrangian multiplier question

So I have to find the min and max values of f(x,y,z) = x^4 + y^4 + z^ 4 given the constraint x^2 + y^2 + z^2 = 1. I've found the points (+-1/sqrt(3),1/sqrt(3) ,1/sqrt(3)), (+-1/sqrt(3),-1/sqrt(3) ,1/sqrt(3)) ... etc all of which have the f-value of 1/3 when x =/= 0 & y =/= 0 & z =/= 0 (this will probably turn out to be the min?)

My issue is the case when one value is zero and the other two are nonzero. I have 4x^3 = 2Lx (I'm using L for lambda here), then 2x^3 - Lx = 0, x(2x^2 - L) = 0....L = +-1/sqrt(2). From there, I have 2y^2 = L = 1/sqrt(2), y^2 = 1/2sqrt(2), and I'm not sure where to go from there for this case.

So I have to find the min and max values of f(x,y,z) = x^4 + y^4 + z^ 4 given the constraint x^2 + y^2 + z^2 = 1. I've found the points (+-1/sqrt(3),1/sqrt(3) ,1/sqrt(3)), (+-1/sqrt(3),-1/sqrt(3) ,1/sqrt(3)) ... etc all of which have the f-value of 1/3 when x =/= 0 & y =/= 0 & z =/= 0 (this will probably turn out to be the min?)

My issue is the case when one value is zero and the other two are nonzero. I have 4x^3 = 2Lx (I'm using L for lambda here), then 2x^3 - Lx = 0, x(2x^2 - L) = 0....L = +-1/sqrt(2). From there, I have 2y^2 = L = 1/sqrt(2), y^2 = 1/2sqrt(2), and I'm not sure where to go from there for this case.

I don't know how you found those points; they do not follow from the Lagrange conditions. For Lagrangian L = x^4 + y^4 + z^4 + u*(x^2 + y^2 + z^2 -1) [using u as the Lagrange multiplier and L as the Lagrangian] the conditions dL/dx = dL/dy = dL/dz = 0 give 2x - 2ux = 0, 2y - 2uy = 0, 2z - 2uz = 0. Thus, when x,y,z =/= 0 we have u = -1, and we have NO other conditions except g = 0. Although the Lagrange multiplier rule does hold at a max or min in this case (because the constraint qualification always holds), it gives no information. Instead, you can look at x1 = x^2, x2 = y^2, x3 = z^2 to get the problem max/min x1^2 + x2^2 + x3^2, subject to x1 + x2 + x3 = 1 and x1, x2, x3 >= 0. Using the Karush-Kuhn-Tucker conditions we find that x1 = x2 = x3 = 1/2 is the min, and that there are three maxima: (x1 = 0, x2 = x3 = 1/2) and permutations of this. Of course, now you can take square roots to get x, y and z.